Union of two bounded sets is bounded Suppose $(X,d)$ is a metric space, for a nonempty subset $A$ of $X$, define
$$ \delta(A) = \sup_{x, y \in A} d(x,y) $$
$A$ is bounded if $\delta(A) < \infty $. If $A$, $B$ are bounded, does it follow that $A \cup B$ is bounded ? I know that $A \cap B$ is bounded since $A \cap B \subseteq B$, and the fact that $A \subseteq B \implies \delta(A) \leq \delta(B) $. But how about the $\cup$ ?
thanks
 A: Let's $a\in A$ and $b\in B$ be fixed. Take any $x\in A , y\in B$, then $$d(x,y) \leq d(x,a ) +d(a,b) +d(b,y) \leq \delta (A) +d(a,b) +\delta (B)$$
thus $A\cup B$ is bounded.
A: And, in order to prove David Mitra's answer, you should:


*

*Use the triangle inequality .

*Convince yourself of the following fact:


$$
\mathrm{sup}_{x\in A,y\in B} \left\{ x + y\right\} \leq \mathrm{sup}_{x\in A} \left\{ x \right\} + \mathrm{sup}_{y\in B} \left\{ y\right\} \ .
$$
EDIT. And the second inequality comes from the fact that, for all $x\in A$ and all $y\in B$, you have
$$
x + y \leq \mathrm{sup}_{x\in A} \left\{ x \right\} + \mathrm{sup}_{y\in B} \left\{ y\right\} \ .
$$
Now, apply $\mathrm{sup}_{x\in A,y\in B}$ to both sides of this inequality and you are done.
Exercise. Why don't you have, in general, an equality 
$$
\mathrm{sup}_{x\in A,y\in B} \left\{ x + y\right\} = \mathrm{sup}_{x\in A} \left\{ x \right\} + \mathrm{sup}_{y\in B} \left\{ y\right\} \quad \text{?}
$$ 
:-)
A: I would reason as follows:
$$\delta(A\cup B)=\sup(\delta(A),\delta(B),\sup_{a\in A, b\in B}d(a,b))$$
Since $A$ and $B$ are bounded, then, by definition, $\delta(A) < \infty$ and $\delta(B) < \infty$. One of the axioms of metric spaces (Kreyszig, 1978, page 3, axiom (M1)) is: "$d$ [the metric] is real-valued, finite and nonnegative". Therefore, $\sup_{a\in A, b \in B}d(a,b) < \infty$, so it follows that $\delta(A \cup B) < \infty$, which means that $A\cup B$ is bounded.
